Question

question 10 of 17
how much heat energy is required to boil 70.2 g of ammonia, nh₃? the molar heat of vaporization of ammonia is 23.4 kj/mol.

Explanation:

Step1: Calculate moles of ammonia

First, find the molar mass of $NH_3$. Nitrogen (N) has a molar - mass of approximately 14.01 g/mol and hydrogen (H) has a molar - mass of approximately 1.01 g/mol. So, the molar mass of $NH_3$ is $M = 14.01+3\times1.01=17.04$ g/mol. The number of moles $n$ of $NH_3$ is calculated by the formula $n=\frac{m}{M}$, where $m = 70.2$ g. So, $n=\frac{70.2}{17.04}$ mol.
$n=\frac{70.2}{17.04}=4.12$ mol

Step2: Calculate heat energy

The heat energy $q$ required for vaporization is given by the formula $q = n\times\Delta H_{vap}$, where $\Delta H_{vap}=23.4$ kJ/mol and $n = 4.12$ mol.
$q=4.12\times23.4$ kJ
$q = 96.4$ kJ

Answer:

96.4 kJ